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| Elections in Ohio |
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The 1836 United States presidential election in Ohio took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.
Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%. Ohio was the home state of William Henry Harrison.
Results
| 1836 United States presidential election in Ohio[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | Francis Granger of New York | 104,958 | 51.87% | 21 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 96,238 | 47.56% | 0 | 0.00% | ||
| N/A | Others | Others | 1,137 | 0.56% | 0 | 0.00% | ||
| Total | 202,333 | 100.00% | 21 | 100.00% | ||||
See also
References
- ^ "1836 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved December 23, 2013.
State and district results of the 1836 United States presidential election | ||
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